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Destructive Interference and coherent light in anti-phase

  1. Mar 24, 2013 #1
    I have a question about coherent light in anti-phase, I'm probably thinking about this wrong but I just cant understand what happens to the energy in the wave when two waves of the same wavelength, frequency and amplitude are in anti-phase as the resultant wave is 0.
     
  2. jcsd
  3. Mar 24, 2013 #2

    sophiecentaur

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    Good question. The energy HAS to go somewhere. In most interference phenomena, it will be directed elsewhere in an interference pattern (e.g. two slits experiment). It may turn up as stored energy in a resonant standing wave pattern or be dissipated in a resistive load / absorbing surface, somewhere.
     
  4. Mar 24, 2013 #3
    Thanks Sophie, I was actually thinking about the double slit experiment. Could you possible expand on this point:
     
  5. Mar 24, 2013 #4

    Dale

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    For two light waves in vacuum, if you have destructive interference in one location then you will have constructive interference in another location. The energh "lost" in the destructive interference region is exactly equal to the energy "gained" in the constructive interference region.
     
  6. Mar 24, 2013 #5
    Say a wave was sent through a waveguide one of length being an integer of the wavelength and the other being a non integer value, this produces destructive interference and a wave with a lower amplitude giving less intensity and energy, in this scenario there are no other waves present where has the energy gone?
     
  7. Mar 24, 2013 #6

    Dale

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    In that case the wave is not travelling through vacuum, it is interacting with the waveguide. The lost energy goes into the material of the waveguide itself, generally heating it, but potentially doing useful work.

    For a wave in vacuum, what I said above holds, which is why I specified "in vacuum" in my comment. For a wave not in vacuum then energy can go to or from matter also.
     
  8. Mar 24, 2013 #7
    Thank you very much Dale, I feel I have a greater understanding now. :smile:
     
  9. Mar 24, 2013 #8

    jtbell

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    In this case the energy is "redistributed" from the locations of the minima to the locations of the maxima. The maxima are "brighter" (contain more energy) than the light would have carried if there had been no interference, just as the minima are "fainter" (carry less energy).

    Specifically, if the light from one slit (by itself) has intensity I0 on the screen, then with two slits the maxima have (maximum) intensity 4I0. The maximum intensity first doubles because there are two slits, then it doubles again because of redistribution of energy from the minima.
     
  10. Mar 24, 2013 #9

    sophiecentaur

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    Some numbers could help. Take an ideal case. At the maximum of the pattern, the amplitude will be 2A, where A would be the amplitude of one contribution. That 4 times the power. In the null, the amplitude is zero, due to desctructive interference. The power has been re-distributed but the same total power will arrive at the screen.
     
  11. Mar 24, 2013 #10
    thanks jtbell and sophie the explanations are very clear and make perfect sense. Is it known how this energy transfers from one wave to another?
     
  12. Mar 24, 2013 #11

    sophiecentaur

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    There is no "transfer" of energy. It's just the result of superposition of the fields at different places. (It's a result of the transit times and phases as the two contributions arrive at each point on the screen) Google "Double slit interference" and the diagrams, explaining it, are all over the place - the Hyperphysics site usually has understandable pages for pretty much every part of Physics at this level.
     
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